All categories

3 years ago

When light reflects from a surface, there is a change in its

1 answer:

6 0

The answer is: none of the above.

Explanation:

When light reflects from a surface, the frequency, wavelength, and speed do not change. They remain the same.

You might be interested in

A 60-kg person is traveling in a car moving at 16 m/s, when the car hits a barrier. The person is not wearing a seat belt, but i

marin [14]

Using conservation of momentum, we can solve for the force that the air bag exerts on the person.

Recall the equation for momentum (p):

$p = mv = F*dt$

We can solve for total momentum, then divide by out time interval. This gets us:

$(60kg)(16m/s) = F(0.2s)


F = 4800N$

F = 4800N

4 0

3 years ago

For light of wavelength 589 nm, calculate the critical angles for the following substances when surrounded by air.

NeX [460]

(1) The critical angles for the fused quartz is 44⁰,

(2) The critical angles for the polystyrene is 39⁰.

(3) The critical angles for the sodium chloride is 41⁰.

<h3>What is Critical angles?</h3>

In optics, the critical angle is the topmost angle at which a shaft of light traveling in one transparent medium can strike the boundary between that medium and another with a lower refractive indicator without being fully reflected within the first

Critical angles for the different medium.

θ $=sin^{-1}\frac{n_{2} }{n_{1} }$

where;

$n_{2}$ is the refractive index of air = 1

$n_{1}$ is the refractive index of the given substances.

Refractive index of quartz glass = 1.46

Refractive index of polystyrene = 1.59

Refractive index of sodium chloride = 1.54

Critical angles for quartz glass:

$\theta_{\text {crit }}=\sin ^{-1} \frac{1.00029}{1.46}\\$

$\theta_{\text {crit }}=44^{o}$

Critical angles for polystyrene

$$\theta_{\text {crit }}=\sin ^{-1} \frac{1.00029}{1.59}\\\\$

$\theta_{\text {crit }}=39^{\circ}$

Critical angles for Sodium Chloride

$$\theta_{\text {crit }}=\sin ^{-1} \frac{1.00029}{1.54}\\$

$\theta_{\text {crit }}=41^{o}$

Thus, the critical angles for the quartz glass is 44⁰, the critical angles for the polystyrene is 39 ⁰ and the critical angles for the sodium chloride is 41⁰.

To know more about critical angle visit:
brainly.com/question/3314727

#SPJ4

I understand that the question you are looking for is:

"For light of wavelength 589 nm, calculate the critical angles for the following substances when surrounded by air.

(a) Fused Quartz

(b) Polystyrene

8 0

2 years ago

Describe the effect of the amplitude on the velocity of the pulse!???

Andre45 [30]

Answer:

Amplitude increases with decreasing velocity.

Explanation:

At the same time, an increase in attention takes place

5 0

3 years ago

Ted wants to hang a wall clock on the wall by using a string. If the mass of the wall clock is 0. 250 kilograms, what should be

antoniya [11.8K]

Sum of all forces = mass * acceleration

Ft= tension force
Fw= force of gravity (Fw= mass* acceleration of gravity which is 9.8 this only applies to force of gravity)

Ft- Fw = 0 (there is no acceleration)
Ft = Fw
Ft= m*g
Ft= 0.250kg*9.8m/s
Ft= 2.45N

7 0

2 years ago

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

3 years ago